# Math Help - Difficult Series Question

1. ## Difficult Series Question

Find the value of b for which

(summation function; n=0, infinity) e^nb = 1 + e^b + e^2b + e^3b + ··· = 9

How do I go about doing this? Thanks!

2. ## Re: Difficult Series Question

Well, this is an infinite geometric series with first term = 1 and ratio = e^b.
There's a formula for that (and maybe even a smartphone app!)...

3. ## Re: Difficult Series Question

Originally Posted by bhaktir
Find the value of b for which

(summation function; n=0, infinity) e^nb = 1 + e^b + e^2b + e^3b + ··· = 9

How do I go about doing this? Thanks!
Hi bhaktir,

$\sum_{n=0}^{\infty}e^{nb}=1 + e^{b} + e^{2b} + e^{3b} +\cdots= 9$

$e^{b} + e^{2b} + e^{3b} + \cdots= 8$

$e^{b}\left(1 + e^{b} + e^{2b} + e^{3b} +\cdots \right)=8$

$e^{b}\sum_{n=0}^{\infty}e^{nb}=8$

$9e^{b}=8~;~\mbox{Since, }\sum_{n=0}^{\infty}e^{nb}=9$

$e^{b}=\frac{8}{9}$

$b=\ln\left(\frac{8}{9}\right)$

4. ## Re: Difficult Series Question

Really helpful! Thank you so much, Sudhraka!!